Chapter 4. Discrete Random Variables
A random variable is a way of recording a quantitative variable of a random experiment.
A variable which can take on only finitely many different values is called discrete.
Example: The number of girls in a family of 8 children
Example: The number of seeds which successfully germinate when 50 seeds are planted
Continuous random variables
• A random variable which can take on any value (ie, all values) in a certain interval is called a continuous random variable.
• EX. The height in centimeters of a 16 year old Canadian male.
• Ex. The dosage in ml. of a certain pain killer
Example
• Let 3 coins be tossed and let x denote the number of heads
• Possible values for x are 0, 1, 2, and 3,
• As done earlier, it is easy to compute
• Pr(0) = 1/8, Pr(3) = 1/8, and Pr(1) = Pr(2) = 3/8
• Notation: We will also use the notation
• P(x = 0) = 1/8, and so on.
Properties of Probability, P( X = xi )
1)(0 (1) ixXP
1)( (2)1
n
iixXP
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Definition
Example
Graph the probability distribution of the random variable obtained by flipping an unbiased coin two times and counting the number of times heads comes up.
Solution
• Possible values of x are 0, 1, 2, and a quick check shows P(0) = ¼, P(1) = 1/2, and P(2) = ¼.
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Probability distribution for a two-coin toss
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Definition
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Definition
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Definition
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Procedure
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Figure 4.6 Shapes of two probability distributions for a discrete random variable x
Example
• Medical research shows a certain type of chemotherapy is successful 70% of the time. Suppose 5 patients are treated and let x denote the number of successes. One can show
• x 0 1 2 3 4 5
P(x) .002 .029 .132 .309 .360 .168
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Graph of p(x)
Find the mean and interpret
• Applying the formulae we obtain
• Mean = 3.50
• Consider a large number of trials, each consisting of treating 5 patients. On average, the number of successes will be 3.5. Thus, if 200 trials each of 5 patients is conducted, we would expect and average of 3.5 successes per trial for a total of 700 successes for 1000 patients
Find the standard deviation and interpret
• Using the formula you can check that• Standard deviation = 1.02.• Using Empirical Rule, would expect that
approximately 68% of times the trial is repeated, the outcome will be between 3.5-1.02 and 3.5+1.02, i.e., will lie in the interval [2.48, 4.52].
• What is the actual percentage of times the outcome will be in that interval?
Binomial Experiment
A binomial experiment is one that:
1) Has a fixed number of trials (n)
2) These trials are independent
3) Each trial must have all outcomes classified into two categories (Success or Failure)
4) The probability of success remains constant for all trials.
Notation:
• S = success and P(S) = p
• F = Failure and P(F) = q = 1- p
• n = fixed number of trials
• x = specific number of successes in n trials
• P(x) = the probability of getting exactly x successes among n trials
Factorials
0! = 1
1! = 1
2! = 2 * 1
3! = 3 * 2 * 1
4! = 4* 3 * 2 * 1
n! = n*(n-1)!
Factorials
0! = 1
1! = 1
2! = 2 * 1=2
3! = 3 * 2 * 1=6
4! = 4* 3 * 2 * 1=24
n! = n*(n-1)!
Binomial Probability Distribution
In a binomial experiment, with constant probability p of success at each trial, the probability of x successes in n trials is given by
xnxqpxxn
nsuccessesxP
!)!(
!) (
ExampleShaq is a basketball player who takes a lot of free throws. The probability of Shaq making a free throw is 0.60 on each throw.
With 3 free throws what is the probability that he makes 2 shots?
Shaq is a basketball player who takes a lot of free throws. The probability of Shaq making a free throw is 0.60 on each throw.
With 3 free throws what is the probability that he makes 2 shots?
0.432
)4(.)6(.!2)!23(
!3)2( 232
xP
Example
Example
Flipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
Example
Flipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
)8(...)3()2()2( xPxPxPxP
Example
Flipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
)1(1
)8(...)3()2()2(
xP
xPxPxPxP
Example
Flipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
)0()1(1
)1(1
)8(...)3()2()2(
xPxP
xP
xPxPxPxP
ExampleFlipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
8071 )6(.)4(.!0 !8
!8)6(.)4(.
!1 !7
!81
)0()1(1
)1(1
)8(...)3()2()2(
xPxP
xP
xPxPxPxP
ExampleFlipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
894.)6(.)4(.!0 !8
!8)6(.)4(.
!1 !7
!81
)0()1(1
)1(1
)8(...)3()2()2(
8071
xPxP
xP
xPxPxPxP
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Table 4.4
How to use the Binomial Tables
•First find the appropriate table for the particular value of n
•then find the value of p in the top row
•Find the row corresponding to k and find the intersection with the column corresponding to the value of p
•The value you obtain is the cumulative probability, that is P(x ≤ k)
•N=10, p = 0.7: P(x ≤ 4) = 0.047
•N=10, p = 0.7: P(x = 4) = P(x ≤ 4) - P(x ≤ 3) = 0.047-0.011=0.036
•N=10, p = 0.7: P(x > 4) = 1- P(x ≤ 4)
= 1 - 0.047 = 0.953
ExampleFlipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
ExampleFlipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
)1(1
)8(...)3()2()2(
xP
xPxPxPxP
ExampleFlipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.
894.
106.01
)1(1
)8(...)3()2()2(
xP
xPxPxPxP
pq
npqnp
1
Mean and Standard deviation
Keys to success
Learn the binomial table.
Be able to recognize binomial distributions and when you do apply the appropriate formulas and tables.